Is every function from a subset of $[0,\infty)$, which is closed under addition and contains $0$, into a Polish space continuous?

Question

Let $I\subseteq[0,\infty)$ be closed under addition with $0\in I$ and $(E,d)$ be a Polish space. Is every function $f:I\to E$ continuous?

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The answer seems to be "yes", if $I\subseteq\mathbb N_0$, since for every $t_0\in I$ $$d\left(f\left(t\right),f\left(t_0\right)\right)=0\;\;\;\text{for all }t\in I\text{ with }|t-t_0|<1\;.$$

But can we show (at least one-sided) continuity for general $I$? Probably not, since the nice property above was $$\inf_{x,y\in I}|x-y|\ge 1\;,$$ but for $I=\mathbb Q\cap[0,\infty)$ $$\inf_{x,y\in I}|x-y|=0\;.$$ Clearly, that's no counter-example (and I'm really bad in finding them), but it shows, that the argumentation above doesn't work in this case.

Answer 1

Added: Answer to new question: No, just take $I = \mathbb {Q}\cap [0,\infty)$ as you did earlier. Define $f:I\to \mathbb {R}$ by setting $f(x) = 1/x, x > 0,f(0)=0.$

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Certainly not: Let $I= \{0\} \cup \{1/n: n \in \mathbb {N}\},$ with the topology it inherits from the usual one on $\mathbb {R}.$ Define $f(1/n) = 1, n = 1,2,\dots , f(0) = 0.$ Then $f$ is discontinuous at $0.$

Answer 2

No, not necessarily, since it depends on the topology on $I$. If $I$ has a trivial topology, where the only open sets are $I$ and $\emptyset$, then it is easy to construct a counterexample. Let $I = \mathbb{Q}$ be the set of rational numbers, but with the trivial topology I just defined, and let $E$ be the set of real numbers, with the usual topology. Then the embedding function $f:I\to E,$ $x\mapsto x$, is not continuous, since the pre-image of every open set is not necessarily open. For example, $f^{-1}((0, 1)) = (0, 1)\cap\mathbb{Q}$ is not open in $I$.

Answer 3

If $I={0,1}$ with topology given by the empty set and the whole set is applied trivially on $I$ with point topology then this application is not continuous. The preimage of the open set $\{1\}$ is not opened.